What is the derivative of x^{4/x} ?

The derivative of x^{4/x} is 4x^{(4-2x)/x}(-ln(x)+1)

What is the first derivative of x^{4/x} ?

The first derivative of x^{4/x} is 4x^{(4-2x)/x}(-ln(x)+1)

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\frac{d}{d x} (x^{\frac{4}{x}})

4 x^{\frac{4 - 2 x}{x}} (- ln (x) + 1)

Solution steps

\frac{d}{d x} (x^{\frac{4}{x}})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{\frac{4}{x}} = e^{\frac{4}{x} l n (x)}

= \frac{d}{d x} (e^{\frac{4}{x} l n (x)})

Apply the chain rule:

e^{\frac{4}{x} l n (x)} \frac{d}{d x} (\frac{4}{x} ln (x))

= e^{\frac{4}{x} l n (x)} \frac{d}{d x} (\frac{4}{x} ln (x))

\frac{d}{d x} (\frac{4}{x} ln (x)) = \frac{4 ( - ln ( x ) + 1 )}{x ^{2}}

= e^{\frac{4}{x} l n (x)} \frac{4 ( - ln ( x ) + 1 )}{x ^{2}}

Simplify

e^{\frac{4}{x} l n (x)} \frac{4 ( - ln ( x ) + 1 )}{x ^{2}} : 4 x^{\frac{4 - 2 x}{x}} (- ln (x) + 1)

= 4 x^{\frac{4 - 2 x}{x}} (- ln (x) + 1)

y^{^{'}} = x (7 - y)

x \to 0 lim (5 (x - 3) ln (x - 3))

\int_{- \infty}^{0} \frac{x}{1 + x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{2}})

What is the derivative of x^{4/x} ?
The derivative of x^{4/x} is 4x^{(4-2x)/x}(-ln(x)+1)
What is the first derivative of x^{4/x} ?
The first derivative of x^{4/x} is 4x^{(4-2x)/x}(-ln(x)+1)